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Three point charges, \(-5.8 \times 10^{-9} \mathrm{C},-9.0 \times 10^{-9} \mathrm{C},\) and \(+7.3 \times 10^{-9} \mathrm{C},\) are fixed at different positions on a circle. The total electric potential at the center of the circle is \(-2100 \mathrm{V}\) . What is the radius of the circle?

Short Answer

Expert verified
The radius of the circle is approximately 32.1 m.

Step by step solution

01

Understand the Electric Potential Formula

The electric potential \( V \) at a point due to a point charge \( q \) at a distance \( r \) is given by the formula \[ V = \frac{kq}{r}\]where \( k \) is Coulomb's constant \( (8.99 \times 10^9 \text{ N m}^2/\text{C}^2) \).
02

Calculate Total Electric Potential

Given three point charges, the total electric potential \( V_{\text{total}} \) at the center of the circle is the sum of the potentials due to each charge:\[V_{\text{total}} = V_1 + V_2 + V_3 = \frac{kq_1}{r} + \frac{kq_2}{r} + \frac{kq_3}{r}\]Thus, \[V_{\text{total}} = \frac{k(q_1 + q_2 + q_3)}{r}\]where \( q_1 = -5.8 \times 10^{-9} \text{ C} \), \( q_2 = -9.0 \times 10^{-9} \text{ C} \), \( q_3 = 7.3 \times 10^{-9} \text{ C} \).
03

Substitute Known Values

Substitute the known values into the equation:\[-2100 = \frac{8.99 \times 10^9 ( -5.8 \times 10^{-9} + -9.0 \times 10^{-9} + 7.3 \times 10^{-9})}{r}\]
04

Simplify the Expression

Calculate the sum of the charges:\[-5.8 \times 10^{-9} + -9.0 \times 10^{-9} + 7.3 \times 10^{-9} = -7.5 \times 10^{-9} \text{ C}\]Now the equation becomes:\[-2100 = \frac{8.99 \times 10^9 \times -7.5 \times 10^{-9}}{r}\]
05

Solve for the Radius

Rearrange the equation to solve for \( r \):\[r = \frac{8.99 \times 10^9 \times 7.5 \times 10^{-9}}{2100}\]Calculate the radius:\[r \approx 32.1 \text{ m}\]
06

Conclusion

The radius of the circle is approximately \( 32.1 \text{ m} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Coulomb's Law
Coulomb's Law is a fundamental principle in physics. It describes the force between two charged objects. The equation is:\[F = \frac{k |q_1 q_2|}{r^2}\]Here, \(F\) is the force between the charges, \(q_1\) and \(q_2\) are the amounts of the charges, \(r\) is the distance separating the charges, and \(k\) is Coulomb's constant. This constant \(k\) has a value of \(8.99 \times 10^9 \text{ N m}^2/\text{C}^2\). This law illustrates that the electric force is inversely proportional to the square of the distance between the charges. This means as the distance \(r\) increases, the force decreases, and vice versa. Understanding this relationship helps in visualizing how charges interact in electric fields, influencing electric potential calculations such as the one given in the problem.
Point Charges
Point charges are a simplified model of charges that assume all of the charge is concentrated at a single point in space. This is an abstraction that makes calculations easier. By treating charges as point charges, we can easily compute the electric fields and potentials they create. Point charges have no dimensions, hence allow us to apply straightforward mathematical models. For example, in the given problem, understanding each charge as a point charge helps when using formulas related to electric potential, like the formula \( V = \frac{kq}{r} \). By considering charges on a circle, we simplify the complex real-world scenario to a problem with clear variables and formulas. This not only aids us in solving exercises but also in understanding electric field interactions conceptually.
Electric Field
The electric field is a vector field around charged particles that represents the force experienced by a small test charge placed within the field. Mathematically, it is defined as:\[E = \frac{F}{q}\]where \(E\) is the electric field, \(F\) is the force experienced by the test charge \(q\). The units for electric field are N/C (newtons per coulomb). An electric field points away from positive charges and towards negative charges, indicating the direction of the force that a positive test charge would experience. Calculating the electric field at certain points helps in understanding the distribution and strength of forces around charges. In exercises like the one provided, the knowledge about electric fields helps predict how charges will influence each other and affect the overall potential at a point, such as the center of the circle in the problem.
Physics Problem Solving
Problem-solving in physics requires a series of logical steps. First, understand the principles involved—such as Coulomb's Law for electric forces. Then, identify all known quantities and represent unknowns clearly. In the provided exercise, the known quantities are the charges and the total electric potential, while the unknown is the radius of the circle. Using formulas, substitute the known values and simplify the equations to isolate the unknown variable. Break down the problem logically: calculate each part step-by-step, checking that each stage makes sense and aligns with physical principles.
  • Break the problem into smaller parts, like calculating total charge influence.
  • Translate word problems into mathematical expressions.
  • Verify your results with units and logical sense.
Approaching problems systematically demystifies complex topics and helps find solutions, just like how we solved for the radius in the exercise.

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Most popular questions from this chapter

A charge of \(-3.00 \mu \mathrm{C}\) is fixed in place. From a horizontal distance of \(0.0450 \mathrm{m},\) a particle of mass \(7.20 \times 10^{-3} \mathrm{kg}\) and charge \(-8.00 \mu \mathrm{C}\) is fired with an initial speed of 65.0 \(\mathrm{m} / \mathrm{s}\) directly toward the fixed charge. How far does the particle travel before its speed is zero?

A capacitor is constructed of two concentric conducting cylindrical shells. The radius of the inner cylindrical shell is \(2.35 \times 10^{-3} \mathrm{m},\) and that of the outer shell is 2.50 \(\times 10^{-3} \mathrm{m}\) . When the cylinders carry equal and opposite charges of magnitude 1.7 \(\times 10^{-10} \mathrm{C}\) , the electric field between the plates has an average magnitude of \(4.2 \times 10^{4} \mathrm{V} / \mathrm{m}\) and is directed radially outward from the inner shell to the outer shell. Determine (a) the magnitude of the potential difference between the cylindrical shells and \((\text { b) the capacitance of this capacitor. }\)

A positive point charge \(\left(q=+7.2 \times 10^{-8} \mathrm{C}\right)\) is surrounded by an equipotential surface \(A,\) which has a radius of \(r_{A}=1.8 \mathrm{m}\) . A positive test charge \(\left(q_{0}=+4.5 \times 10^{-11} \mathrm{C}\right)\) moves from surface \(A\) to another equipotential surface \(B,\) which has a radius \(r_{B} .\) The work done as the test charge moves from surface \(A\) to surface \(B\) is \(W_{A B}=-8.1 \times 10^{-9} \mathrm{J} .\) Find \(r_{B}\)

What is the capacitance of a capacitor that stores 4.3\(\mu \mathrm{C}\) of charge on its plates when a voltage of 1.5 \(\mathrm{V}\) is applied between them?

Two charges \(A\) and \(B\) are fixed in place, at different distances from a certain spot. At this spot the potentials due to the two charges are equal. Charge \(A\) is 0.18 m from the spot, while charge \(B\) is 0.43 m from it. Find the ratio \(q_{B} / q_{A}\) of the charges.

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